Electrical Machines I - Lecture 4: Equivalent Circuits Transformers

Impedance Ratio

• Impedance ratio ($Z<em>2/Z</em>1$) is equal to the square of the voltage transformation ratio.

• An impedance $Z<em>2$ in the secondary becomes $Z</em>2/K^2$ when transferred to the primary.

• An impedance $Z<em>1$ in the primary becomes $K^2*K</em>1$ when transferred to the secondary.

• Similarly, $R<em>2/R</em>1 = K^2$ and $X<em>2/X</em>1 = K^2$

Transferring Parameters

• A resistance $R<em>1$ in the primary becomes $K^2*R</em>1$ when transferred to the secondary.

• A resistance $R<em>2$ in the secondary becomes $R</em>2/K^2$ when transferred to the primary.

• A reactance $X<em>1$ in the primary becomes $K^2*X</em>1$ when transferred to the secondary.

• A reactance $X<em>2$ in the secondary becomes $X</em>2/K^2$ when transferred to the primary.

*When transferring impedance (R or X) from primary to secondary, multiply it by $K^2$
*When transferring impedance (R or X) from secondary to primary, divide it by $K^2$
*When transferring voltage or current from one winding to the other, only K is used.

Shifting Impedances in a Transformer

• Resistances and reactances external to the windings.

• Resistance and reactance of one winding can be transferred to the other by appropriately using the factor $K^2$.

• Transformer can be:

  • Referred to primary

  • Referred to secondary

Equivalent Resistances

• Individual resistances

• Referred to primary

• Referred to secondary

Equivalent Resistance

• Resistance of the two windings can be transferred to either the primary or secondary side without affecting the transformer's performance.

• Transferring resistances to one side simplifies calculations.

Total Copper Loss

• Total copper loss = $I^2 R<em>{1e} = I</em>1^2 R<em>1 + I</em>2^2 R_2$

• Equivalent resistance $R_{1e}$ simplifies calculations by allowing computation on one side only.

• Similarly, it is possible to refer the equivalent resistance to the secondary winding.

• Total copper loss $= I<em>1^2[R</em>1 + \frac{R_2}{K^2}]$

• $R<em>1' = K^2 R</em>1$ (primary resistance referred to secondary)

Transformer Referred to Primary

• When secondary impedance is transferred to the primary, it is divided by $K^2$.

• Equivalent secondary resistance/reactance referred to primary is denoted by $R<em>2'$ or $X</em>2'$.

• Equivalent resistance of transformer referred to primary: $R<em>{01} = R</em>1 + R<em>2' = R</em>1 + \frac{R_2}{K^2}$

• Equivalent reactance of transformer referred to primary: $X<em>{01} = X</em>1 + X<em>2' = X</em>1 + \frac{X_2}{K^2}$

• Equivalent impedance of transformer referred to primary: $Z<em>{01} = \sqrt{R</em>{01}^2 + X_{01}^2}$

• Note that secondary now has no resistance or reactance.

Transformer Referred to Secondary

• When primary impedance is transferred to the secondary, it is multiplied by $K^2$.

• Equivalent primary resistance/reactance referred to secondary is denoted by $R<em>1'$ or $X</em>1'$.

• Equivalent resistance of transformer referred to secondary: $R<em>{02} = R</em>2 + R<em>1' = R</em>2 + K^2 R_1$

• Equivalent reactance of transformer referred to secondary: $X<em>{02} = X</em>2 + X<em>1' = X</em>2 + K^2 X_1$

• Equivalent impedance of transformer referred to secondary: $Z<em>{02} = \sqrt{R</em>{02}^2 + X_{02}^2}$

• Note that primary now has no resistance or reactance.

Impedance Equivalent

• Transformer primary has resistance $R<em>1$ and reactance $X</em>1$.

• Transformer secondary has resistance $R<em>2$ and reactance $X</em>2$.

• Total impedance of primary winding: $Z<em>1 = R</em>1 + jX_1$

• Total impedance of secondary winding: $Z<em>2 = R</em>2 + jX_2$

• $\frac{R<em>1}{Z</em>1} = \frac{R<em>e}{Z</em>{1e}}$

• $Z<em>{1e} = \sqrt{R</em>e^2 + X_e^2}$

• $Z<em>{2e} = \frac{R</em>2}{K^2} + Z<em>1' = Z</em>2 + K^2 Z_1$

Practice Question 1 Solution

• Given: 33 kVA, 2200/220V, 50Hz single-phase transformer, $R<em>1 = 2.4 \Omega$, $X</em>1 = 6 \Omega$, $R<em>2 = 0.03 \Omega$, $X</em>2 = 0.07 \Omega$

• Transformation ratio: $K = \frac{V<em>2}{V</em>1} = \frac{220}{2200} = 0.1$

• Transformer resistance referred to primary side: $R<em>{ep} = R</em>1 + \frac{R_2}{K^2} = 2.4 + \frac{0.03}{(0.1)^2} = 5.4 \Omega$

• Transformer reactance referred to primary side: $X<em>{ep} = X</em>1 + \frac{X_2}{K^2} = 6 + \frac{0.07}{(0.1)^2} = 13 \Omega$

• Transformer resistance referred to secondary side: $R<em>{es} = R</em>2 + R_1 K^2 = 0.03 + 2.4 (0.1)^2 = 0.054 \Omega$

• Transformer reactance referred to secondary side: $X<em>{es} = X</em>2 + X_1 K^2 = 0.07 + 6 (0.1)^2 = 0.13 \Omega$

Practice Question 2 Solution

• Single-phase transformer, voltage ratio 2500/250V, $R<em>1 = 1.8 \Omega$, $X</em>1 = 4.2 \Omega$, $R<em>2 = 0.02 \Omega$, $X</em>2 = 0.045 \Omega$

• Transformation ratio: $K = \frac{250}{2500} = 0.1$

• Total resistance referred to secondary side: $R<em>{es} = R</em>2 + R_1 K^2 = 0.02 + 1.8 (0.1)^2 = 0.038 \Omega$

• Total reactance referred to secondary side: $X<em>{es} = X</em>2 + X_1 K^2 = 0.045 + 4.2 (0.1)^2 = 0.087 \Omega$

• Impedance of transformer referred to secondary side: $Z<em>{es} = \sqrt{R</em>{es}^2 + X_{es}^2} = \sqrt{(0.038)^2 + (0.087)^2} = 0.095 \Omega$

Practice Questions

1. A 10 kVA 2000/400 V single phase transformer has $R<em>1 = 5 \Omega$, $X</em>1 = 12 \Omega$, $R<em>2 = 0.2 \Omega$ and $X</em>2 = 0.48 \Omega$. Determine the equivalent impedance of the transformer referred to (i) primary side and (ii) secondary side [$Z{01} = 26 \Omega$], [$Z{02} = 1.04 \Omega$].

2. A 100 kVA, 2200/440 V transformer has $R<em>1 = 0.3 \Omega$, $X</em>1 = 1.1 \Omega$, $R<em>2 = 0.01 \Omega$ and $X</em>2 = 0.035 \Omega$.
   Calculate:

   • the equivalent impedance of the transformer referred to the primary

   • total copper losses [$R{01} = 0.55$ $X{01} = 1.975$ $Z_{01} = 2.05$ $P = 1136$ W]

3. A 50-kVA, 4,400/220-V transformer has $R<em>1 = 3.45 \Omega$, $X</em>1 = 5.2 \Omega$, $R<em>2 = 0.009 \Omega$ and $X</em>2 = 0.015 \Omega$.
   Calculate:

   • equivalent resistance as referred to primary [7.05]

   • equivalent resistance as referred to secondary [0.0176]

   • equivalent reactance as referred to both primary and secondary [$X{01} = 11.2$]; [$X{01} = 0.028$]

   • equivalent impedance as referred to both primary and secondary [$Z{01} = 13.23$]; [$Z{01} = 0.03311$]

Equivalent Circuit of a Transformer

• The equivalent circuit of a machine means the combination of fixed and variable resistances and reactances, which exactly simulates performance and working of the machine.

• For a transformer, no load primary current $I_0$ has two components

• $I<em>m = I</em>0 sin \theta_0 = magnetising component$

• $I<em>c = I</em>0 cos \theta_0 = active component$

• $I<em>m$ produces the flux and is assumed to flow through reactance $X</em>0$ called no load reactance while $I<em>c$ is active component representing core losses hence is assumed to flow through the resistance $R</em>0$. Hence equivalent circuit on no load can be shown as in the Fig. This circuit consisting of $R<em>0$ and $X</em>0$ in parallel is called exciting circuit. From the equivalent circuit we can write,

• $\frac{V<em>1}{R</em>0}$

• $\frac{V<em>1}{X</em>0}$

• When the load is connected to the transformer then secondary current $I<em>2$ flows. This causes voltage drop across $R</em>2$ and $X<em>2$. Due to $I</em>2$, primary draws an additional current $I<em>2' = \frac{I</em>2}{K}$. Now $I<em>1$ is the phasor addition of $I</em>0$ and $I<em>2'$. This $I</em>1$ causes the voltage drop across primary resistance $R<em>1$ and reactance $X</em>1$.

Equivalent Circuit Importance

• Important to describe the behavior of an electrical machine in terms of its equivalent circuit.

• Analysis can be done using normal methods of circuit theory.

• $R<em>1$ is the primary winding resistance and $R</em>2$ is the secondary winding resistance.

• $X<em>1$ is the leakage reactance of primary winding and $X</em>2$ is the leakage reactance of the secondary winding.

• The parallel circuit $R<em>0 - X</em>0$ is the no-load equivalent circuit of the transformer.

• The resistance $R<em>0$ represents the core losses (hysteresis and eddy current losses) so that current $I</em>W$ which supplies the core losses is shown passing through $R_0$.

• The inductive reactance $X<em>0$ represents a loss-free coil which passes the magnetizing current $I</em>m$.

• The phasor sum of $I<em>W$ and $I</em>m$ is the no-load current $I_0$ of the transformer

• Note that equivalent circuit has created two normal electrical circuits separated only by an ideal transformer whose function is to change values according to the equation:

Equivalent Circuit Points

i. When the transformer is on no-load, there is no current in the secondary winding. However, the primary draws a small no-load current $I<em>0$. The no-load primary current $I</em>0$ is composed of (a) magnetizing current ($I<em>m$) to create magnetic flux in the core and (b) the current $I</em>W$ required to supply the core losses.
ii. When the secondary circuit of a transformer is closed through some external load $Z<em>L$, the voltage $E</em>2$ induced in the secondary by mutual flux will produce a secondary current $I<em>2$. There will be $I</em>2 R<em>2$ and $I</em>2 X<em>2$ drops in the secondary winding so that load voltage $V</em>2$ will be less than $E_2$

Points to Note

(iii) When the transformer is loaded to carry the secondary current $I_2$, the primary current consists of two components:

(a) The no-load current $I_0$ to provide magnetizing current and the current required to supply the core losses.

(b) The primary current $I<em>2' = K I</em>2$ required to supply the load connected to the secondary.

$\therefore$ Total primary current $I<em>1 = I</em>0 + (-KI_2)$

(iv) Since the transformer in Fig. is now ideal, the primary induced voltage $E<em>1$ can be calculated from the relation: $\frac{E</em>1}{E<em>2} = \frac{N</em>1}{N_2}$

If we add $I<em>1R</em>1$ and $I<em>1X</em>1$ drops to $E<em>1$, we get the primary input voltage $V</em>1$
$V<em>1 = -E</em>1 + I<em>1 (R</em>1 + j X<em>1) = -E</em>1 +I<em>1Z</em>1$

Equivalent Circuit Referred to Primary

If all the secondary quantities are referred to the primary, we get the equivalent circuit of the transformer referred to the primary.

Note that when secondary quantities are referred to primary, resistances/reactances/impedances are divided by $K^2$, voltages are divided by K and currents are multiplied by K.

$R'<em>2=\frac{R</em>2}{K^2}; X'<em>2=\frac{X</em>2}{K^2}; Z'<em>L = \frac{Z</em>L}{K^2}; V'<em>2 = \frac{V</em>2}{K}; I'<em>2=K I</em>2$

$Z<em>{01} = R</em>{01} +jX<em>{01}$ where $R</em>{01} = R<em>1 +R'</em>2; X<em>{01} =X</em>1 +X'_2$

Equivalent circuit referred to primary

• The no-load current $I_0$ in a transformer is only 1-3% of the rated primary current and may be neglected without any serious error.

Equivalent circuit of transformer referred to primary

If all the secondary quantities are referred to the primary, we get the
equivalent circuit of the transformer referred to primary as shown.

When secondary quantities are referred to primary,
resistances/reactances are divided by $K^2$, voltages are divided by K
and currents are multiplied by K.

The equivalent circuit shown is an electrical circuit and can be solved
for various currents and voltages
Thus if $V'<em>2$ and $I'</em>2$ are obtained, then:

• Actual secondary voltage, $V<em>2 = KV'</em>2$

• Actual secondary current, $I<em>2 = \frac{I'</em>2}{K}$

Equivalent circuit of transformer referred to secondary

If all the primary quantities are referred to secondary, we get the equivalent
circuit of the transformer referred to secondary .
Note that when primary quantities are referred
to secondary resistances/reactances/impedances are multiplied by $K^2$, voltages
are multiplied by K. and currents are divided by K.
$I'<em>1 = I</em>1K$
$R'<em>1 = K^2R</em>1; X'<em>1 =K^2X</em>1$
$V'<em>2=K V</em>1; I'<em>1 = \frac{I</em>1}{K}$
$Z<em>{02} = R</em>{02} +jX<em>{02}$ where $R</em>{02} =R<em>2+R'</em>1; X<em>{02} = X</em>2 +X'_1$

Equivalent circuit of transformer referred to secondary

Note: The same final answers will be obtained whether we use the equivalent circuit referred to primary or secondary. The use of a particular equivalent circuit would depend upon the conditions of the problem.

Approximate Voltage Drop in a Transformer

Consider the approximate equivalent circuit of transformer referred to
secondary shown;

At no-load, the secondary voltage is $KV_1$.

When a load having a lagging p.f. $cos \theta<em>2$ is applied, the secondary carries a current $I</em>2$ and voltage drops occur in $(R<em>2 + K^2 R</em>1)$ and
$(X<em>2+K^2X</em>1)$

Consequently, the secondary voltage falls from $KV<em>1$ to $V</em>2$.

Referring to the Figure, we have,

$V<em>2 = KV</em>1 - I<em>2[(R</em>2 + K^2R<em>1) + j(X</em>2 + K^2X_1)]$

$= KV<em>1 - I</em>2(R<em>{02} + jX</em>{02}) = KV<em>1 - I</em>2Z_{02}$

Approximate Voltage Drop in a Transformer

Drop in secondary voltage = $I<em>2Z</em>{02} = KV<em>1 - V</em>2$
$= I<em>2 R</em>{02} cos \theta<em>2 +I</em>2X<em>{02} sin \theta</em>2$

For a load having a leading p.f. $cos \theta_2$, we have,

Approximate voltage drop
$= I<em>2 R</em>{02} cos \theta<em>2 \pm I</em>2X<em>{02} sin \theta</em>2$

Note: If the circuit is referred to primary, then it can be easily
established that:

Approximate voltage drop
$= I<em>1 R</em>{01} cos \theta<em>1 \pm I</em>1X<em>{01} sin \theta</em>1$

Voltage Regulation

The voltage regulation of a transformer is the arithmetic difference (not
phasor difference) between the no-load secondary voltage $(0 V<em>2)$ and the secondary voltage $V</em>2$ on load expressed as percentage of no-load
voltage i.e.

$\% \text{age voltage regulation} = \frac{0 V<em>2 - V</em>2}{0 V_2}$

where:

• $0V<em>2$ = No-load secondary voltage = $KV</em>1$.

• $V_2$ = Secondary voltage on load

As discussed above (voltage drop)
$0 V<em>2 - V</em>2 = I<em>2 R</em>{02} COS \theta<em>2 \pm I</em>2X<em>{02} sin \theta</em>2$

The +ve sign is for lagging p.f. and -ve sign for leading p.f.
The %age voltage regulation is the same whether primary or secondary
side is considered.

Practice Question 1

A 250/125V, 5 kVA single phase transformer has primary resistance of 0.2 Ω and reactance of 0.75Ω. The secondary resistance is 0.05 Ω and reactance of 0.2 Ω. Determine:
i. Its regulation while supplying full load on 0.8 leading p.f
ii. The secondary terminal voltage on full load and 0.8 leading p.f.

Practice Question 2

A 40 kVA, 6600/250 V, 50 Hz transformer is having total reactance of 35 Ω when referred to primary side whereas its primary and secondary winding resistance is 10 Ω and 0.02 Ω, respectively. Find full load regulation of at a p.f. 0.8 lagging.

Solution Q1

$R<em>1 = 0.2 \Omega$, $X</em>1 = 0.75 \Omega$, $R<em>2 = 0.05 \Omega$, $X</em>2 = 0.2 \Omega$, $cos \phi = 0.8 \text{ leading}$

$K = \frac{E<em>2}{E</em>1} = \frac{125}{250} = \frac{1}{2} = 0.5$

(I) $F.L. = \frac{KVA}{V_2} = \frac{5 \times 10^3}{125} = 40 A$

$R<em>{2e} = R</em>2 + K^2R_1 = 0.05 + (0.5)^2 \times 0.2 = 0.1 \Omega$

$X<em>{2e} = X</em>2 + K^2X_1 = 0.2 + (0.5)^2 \times 0.75 = 0.3875 \Omega$

i) Regulation on full load, $cos \phi = 0.8 \text{ leading}$

$sin \phi = 0.6$

$\% R = \frac{I<em>2 R</em>{2e} cos \phi - I<em>2 X</em>{2e} sin \phi}{V_2} \times 100$

$I_2 = \text{Full load current}$

$\% R = \frac{(40 \times 0.1 \times 0.8 - 40 \times 0.3875 \times 0.6)}{125} \times 100 = - 4.88\%$

ii) For secondary terminal voltage, use basic expression of voltage drop.

On no load, $E_2 = 125 V$

$E<em>2 - V</em>2 = I<em>2 [R</em>{2e} cos \phi - X<em>{2e} sin \phi]$ $=40 [0.1 \times 0.8 - 0.3875 \times 0.6] = - 6.1 V$ $V</em>2 = E<em>2 - [- 6.1] = 125 + 6.1$ = 131.1 V \text{ for leading p.f. } E2 < V_2

Solution Q1

Rating of transformer, = 40 kVA = $40 \times 10^3 VA$

Primary resistance, $R_1 = 10 \Omega$

Transformation ratio, $K = \frac{250}{6600} = 0.03788$

Secondary resistance, $R_2 = 0.02 \Omega$

Total resistance, referred to primary side,

$R<em>{ep} = R</em>1 + \frac{R_2}{K^2} = 10 + \frac{0.02}{(0.03788)^2} = 23.94 \Omega$

Total reactance referred to primary side, $X_{ep} = 35 \Omega$

where,

$V<em>1 = \sqrt{(E</em>1 cos \phi + I<em>1 R</em>{ep})^2 + (E<em>1 sin \phi + I</em>1 X_{ep})^2}$

$I_1 = \frac{40 \times 10^3}{6600} = 6.06 A$

$cos \phi = 0.8; sin \phi = sin cos^{-1} 0.8 = 0.6$

$V_1 = \sqrt{(6600 \times 0.8 + 6.06 \times 23.94)^2 + (6600 \times 0.6 + 6.06 \times 35)^2} = 6843.7 V$

$\% Reg = \frac{V<em>1 - E</em>1}{V_1} \times 100 = \frac{6843.7-6600}{6843.7} \times 100 = 3.56\%$

Practice Question

Qn 1.: A single phase transformer with a ratio 5: 1 has primary resistance of 0.4 ohm and reactance of 1.2ohm and the secondary resistance of 0.01 and reactance of 0.04 ohm. Determine the percentage regulation when delivering 125 A at 600 V at
(i) 0.8 p.f. lagging
(ii) 0.8 p.f. leading.

Qn 2.: If the ohmic loss of a transformer is 1% of the output and its reactance drop is 5% of the voltage, determine its regulation when the power factor is
(i) 0.8 lagging
(ii) 0.8 leading
(iii) unity.